Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int A[100010][20];
int L[100010];
vector<int> adjList[100010];
vector<int> vals[100010][20];
vector<int> people[100010];
vector<int> mergeret(vector<int> &v1, vector<int> &v2) {
  vector<int> ret;
  int idx1 = 0, idx2 = 0;
  while (idx1 < v1.size() && idx2 < v2.size()) {
    if (v1[idx1] < v2[idx2])
      ret.push_back(v1[idx1++]);
    else if (v1[idx1] > v2[idx2])
      ret.push_back(v2[idx2++]);
    else {
      ret.push_back(v2[idx2++]);
      idx1++;
    }
  }
  while (idx1 < v1.size()) ret.push_back(v1[idx1++]);
  while (idx2 < v2.size()) ret.push_back(v2[idx2++]);
  while (ret.size() > 10) ret.pop_back();
  return ret;
}
void dfs(int u, int p, int l) {
  L[u] = l;
  A[u][0] = p;
  if (p != -1) vals[u][0] = mergeret(people[u], people[p]);
  for (auto v : adjList[u])
    if (v != p) dfs(v, u, l + 1);
}
vector<int> lca(int p, int q) {
  if (p == q) return people[p];
  if (L[p] < L[q]) swap(p, q);
  vector<int> vl = vals[p][0];
  for (int i = 19; i >= 0; i--)
    if ((L[p] - (1 << i)) >= L[q]) {
      vl = mergeret(vl, vals[p][i]);
      p = A[p][i];
    }
  if (p == q) return vl;
  vector<int> vr = vals[q][0];
  for (int i = 19; i >= 0; i--)
    if (A[p][i] != -1 && A[p][i] != A[q][i]) {
      vl = mergeret(vl, vals[p][i]);
      vr = mergeret(vr, vals[q][i]);
      p = A[p][i];
      q = A[q][i];
    }
  vector<int> ret = mergeret(vl, vr);
  return mergeret(people[A[p][0]], ret);
}
int main() {
  int N, M, Q;
  scanf("%d %d %d", &N, &M, &Q);
  for (int i = 1; i < N; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    u--;
    v--;
    adjList[u].push_back(v);
    adjList[v].push_back(u);
  }
  for (int i = 0; i < M; i++) {
    int x;
    scanf("%d", &x);
    x--;
    people[x].push_back(i + 1);
  }
  for (int i = 0; i < N; i++) {
    sort(people[i].begin(), people[i].end());
    while (people[i].size() > 10) people[i].pop_back();
  }
  dfs(0, -1, 1);
  for (int j = 1; j < 20; j++)
    for (int i = 0; i < N; i++) {
      if (A[i][j - 1] == -1) {
        A[i][j] = -1;
        continue;
      }
      A[i][j] = A[A[i][j - 1]][j - 1];
      if (A[i][j] != -1)
        vals[i][j] = mergeret(vals[i][j - 1], vals[A[i][j - 1]][j - 1]);
    }
  for (int i = 0; i < Q; i++) {
    int u, v, a;
    scanf("%d %d %d", &u, &v, &a);
    u--;
    v--;
    vector<int> ret = lca(u, v);
    a = min(a, (int)ret.size());
    printf("%d ", a);
    for (int i = 0; i < a; i++) printf("%d ", ret[i]);
    printf("\n");
  }
  return 0;
}